Question

use modified euler 's method to find the approximate solution of IVP y' =2xy, y(1)=1, at x=1.5 with h=0.1. if the exact solution is y(x) =ex2-1, find the error.

Answer 1

Question #74942 - Math - Quantitative Methods

use modified euler's method to find the approximate solution of IVP $y' = 2xy$ , $y(1) = 1$ , at $x = 1.5$ with $h = 0.1$ . If the exact solution is $y(x) = ex2 - 1$ , find the error.

**Solution:** for the numerical solution of the differential equation:

$$y ^ {\prime} = f (x, y), \qquad f (x, y) = 2 \cdot x \cdot y$$

use the explicit midpoint method also known as the modified Euler method:

$$y _ {n + 1} = y _ {n} + h \cdot f \left(x _ {n} + \frac {h}{2}, y _ {n} + \frac {h}{2} \cdot f \left(x _ {n}, y _ {n}\right)\right), \quad h = 0. 1, \quad x _ {0} = 1, \quad y _ {0} = 1$$

to find the approximate solution of IVP at $x = 1.5$ , it takes 5 steps:

$$\begin{array}{l} x _ {1} = x _ {0} + h = 1. 1, y _ {1} = y _ {0} + h \cdot f \left(x _ {0} + \frac {h}{2}, y _ {0} + \frac {h}{2} \cdot f \left(x _ {0}, y _ {0}\right)\right) \\ = 1 + 0. 1 \cdot f (1. 0 5, 1 + 0. 0 5 \cdot f (1, 1)) = 1 + 0. 1 \cdot f (1. 0 5, 1. 1) = 1. 2 3 1 0 \\ \end{array}$$ $$\begin{array}{l} x _ {2} = x _ {1} + h = 1. 2, y _ {2} = y _ {1} + h \cdot f \left(x _ {1} + \frac {h}{2}, y _ {1} + \frac {h}{2} \cdot f \left(x _ {1}, y _ {1}\right)\right) \\ = 1. 2 3 1 + 0. 1 \cdot f (1. 1 5, 1. 2 3 1 + 0. 0 5 \cdot f (1. 1, 1. 2 3 1)) \\ = 1. 2 3 1 + 0. 1 \cdot f (1. 1 5, 1. 3 6 6 4) = 1. 5 4 5 3 \\ \end{array}$$ $$\begin{array}{l} x _ {3} = x _ {2} + h = 1. 3, y _ {3} = y _ {2} + h \cdot f \left(x _ {2} + \frac {h}{2}, y _ {2} + \frac {h}{2} \cdot f \left(x _ {2}, y _ {2}\right)\right) \\ = 1. 5 4 5 3 + 0. 1 \cdot f (1. 2 5, 1. 5 4 5 3 + 0. 0 5 \cdot f (1. 2, 1. 5 4 5 3)) \\ = 1. 5 4 5 3 + 0. 1 \cdot f (1. 2 5, 1. 7 3 0 7) = 1. 9 7 7 9 5 \\ \end{array}$$ $$\begin{array}{l} x _ {4} = x _ {3} + h = 1. 4, y _ {4} = y _ {3} + h \cdot f \left(x _ {3} + \frac {h}{2}, y _ {3} + \frac {h}{2} \cdot f \left(x _ {3}, y _ {3}\right)\right) \\ = 1. 9 7 7 9 5 + 0. 1 \cdot f (1. 3 5, 1. 9 7 7 9 5 + 0. 0 5 \cdot f (1. 3, 1. 9 7 7 9 5)) \\ = 1. 9 7 7 9 5 + 0. 1 \cdot f (1. 3 5, 2. 2 3 5 1) = 2. 5 8 1 4 \\ \end{array}$$ $$\begin{array}{l} x _ {5} = x _ {4} + h = 1. 5, y _ {5} = y _ {4} + h \cdot f \left(x _ {4} + \frac {h}{2}, y _ {4} + \frac {h}{2} \cdot f \left(x _ {4}, y _ {4}\right)\right) \\ = 2. 5 8 1 4 + 0. 1 \cdot f (1. 4 5, 2. 5 8 1 4 + 0. 0 5 \cdot f (1. 4, 2. 5 8 1 4)) \\ = 2. 5 8 1 4 + 0. 1 \cdot f (1. 4 5, 2. 9 4 2 8) = 3. 4 3 4 8 \\ \end{array}$$

exact solution:

$$y (x) = e ^ {x ^ {2} - 1}, y (1. 5) = 3. 4 9 0 3 \quad e r r o r = 3. 4 9 0 3 - 3. 4 3 4 8 = 0. 0 5 5 5 (\sim 1. 6 \%)$$

Answer: $y_{5} = 3.4348$ , error $= 0.0555$ ( $\sim 1.6\%$ )

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